Product of central binomial coefficients I have a question about an equality involving products of central binomial coefficients. If $x_1,...,x_n$ and $y_1,...,y_n$ are positive integers, with $\sum_i x_i = \sum_i y_i$ and 
$$ \binom{2x_1}{x_1} \cdots \binom{2x_n}{x_n} = \binom{2y_1}{y_1}\cdots \binom{2y_n}{y_n}\,, $$
what are the restrictions on the $x_i$ and $y_i$, and is there any solution other than the trivial one $\{x_1,...,x_n\}=\{y_1,...,y_n\}$?
 A: Yes, there are nontrivial solutions.  The first I found, with $n=5$, has
$$
\lbrace x_i \rbrace = \lbrace 2, 5, 8, 13, 19 \rbrace,
\phantom{\infty}
\lbrace y_i \rbrace = \lbrace 3, 4, 6, 14, 20 \rbrace,
$$
with $\sum_i x_i = \sum_i y_i = 47$ and
$$
\prod_{i=1}^5 {2x_i \choose x_i} = \prod_{i=1}^5 {2y_i \choose y_i}
 = 7153522697506948963200000
 = 2^{10} 3^6 5^5 7^3 11^2 13^1 17^1 19^1 23^2 29^1 31^1 37^1.
$$
This is a "list anagram" problem in multiplicative disguise; 
for another such example see puzzle #12
here
and the notes and links in the
solution page.
[EDIT] I see that the OP did not specify that the $x_i$ and $y_i$
be distinct, which makes it a simpler problem (only linear algebra
over the rationals, no lattice reduction).  But the version with
distinct variables is more appealing.

[added later] to spell it out: let $C(m) := {2m \choose m}$,
and suppose all $x_i,y_i$ are at most $M$.
For $1 \leq m \leq M$ let $a_m = \#\{i: x_i=m\} - \#\{i: y_i=m\}$.
Then $x_i,y_i$ are a solution iff
$\sum_{m=1}^M a_m = \sum_{m=1}^M m a_m = 0$ and $\prod_{m=1}^M C(m)^{a_m}=1$.
By unique factorization, the last condition is equivalent to
$\sum_{m=1}^M v_p(C(m)) a_m = 0$ for all primes $p$
(where $v_p$ is the $p$-valuation), and we need only consider $p \leq 2M$.
So we have $\pi(2M)+2$ linear equations in $M$ variables.
There are nontrivial solutions provided $\pi(2M)+2 < M$,
which happens once $M$ is large enough because $\pi(x) = o(x)$;
it turns out that any $M>10$ is large enough.
The equations have integer coefficients, so any solution is rational,
and can be made integral by multiplying by a common denominator.
For example, the 1-dimensional space of solutions for $M=11$
is generated by $(a_1,a_2,\ldots,a_{11}) = (1,-2,0,0,2,-1,0,0,0,1,-1)$,
giving $n=4$, $x_i=1,5,5,10$, and $y_i=2,2,6,11$ with a common sum of $21$
and
$$
\prod_i C(x_i) = \prod_i C(y_i) = 23465490048 = 2^7 3^4 7^2 11^1 13^1 17^1 19^1.
$$
The $x_i$ and $y_i$ are distinct iff the $a_m$ are all $0$ or $\pm 1$.
There's no easy criterion for this, but it's correlated with small norm
$\sum_{m=1}^M |a_m|^2$, so we'll often find an example by applying LLL
(or some other technique for lattice basis reduction)
to the lattice of integer solutions of our linear system equation.
Fortunately the matkerint function of gp does this automatically,
and then vecmax(abs(...)) can be used to detect vectors with all
coordinates $0$ or $\pm 1$.  A bit more work finds such a vector
already for $M=18$, with $n=6$ and
$x_i,y_i = \lbrace 3, 6, 8, 11, 13, 17\rbrace,
\lbrace 4, 5, 7, 10, 14, 18 \rbrace$.
